Thank you for the opportunity to help you with your question!
The first thing we need to do is find out the molecular weight of MgSO4. When we add it up from the periodic table, we arrive at 120.37 g/mol. To see how many mols are left over at the end, we can do a quick calculation of: 1.753g/ 120.37g/mol =.0145 mols of MgSO4.
Too see how much weight the water contributed, all we do is subtract the amount of MgSO4 from the total.
Weight of H20= 3.590g - 1.753g = 1.837g of water
Next, we need to find out how many mols of water are in 1.837g. We do this by dividing it by its molecular weight: 1.837 g/ 18.02 g/mol = .1019 mols of water
lastly, in order to find the the coefficient that goes in front of H20, we divide by the smallest number:
x= .1019/.0145 = 7.030
Please let me know if you need any clarification. I'm always happy to answer your questions.
If it were me, I would just put 7 because x in this case is a coefficient in the formula, as you said.
All sig digs means is that if you were given 3 decimal places in the question, you can have that in your answer. If it is work for your teacher and he/she is strict about it, I would show my work and write: x = 7.030 , x = 7
if 7 is wrong (which i hope it wouldn't be!), I would go through the steps again and keep an extra decimal place while I'm doing the work to double check that it's 7.030, and not 7.031, for example.
I would try 7 though. - because even though you're allowed 3 sig digs in this question, it acts as a coefficient. that's my opinion...
ok, hi again.
I did it again with doing 3 sig digs for the MM instead of 2. (ie, 18.015 for water instead of 18.02, and 120.366 for MgSO4 instead of 120.37)
with that, I got .01456 mols of MgSO4, and .10197 of water. when i divided those, I got 7.003 .
(I'm not sure if your teacher wants you to do 3 sig digs the whole way through though - as in , .102/.015 = 6.8 - but that might be rounding too much.)
DId you get the same answer as me?
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